Tell me if this arguement is right...... I constructed a 1by 1 square on the coordinate plane vertices (0,0) ,(0,1),(1,0),(1,1). Then I said that to graph any continuous function , the graph must touch or cut the square's diagonal , that is y=x.

Apply MVT.
According to MVT, there exists a point c such that, f'(c)={f(1)-f(0)}/(1-0)=1-0=1
Now dy/dx=1. So you can find y.
now substitute a point in the domain to get the value of integration constant.